 Hello and welcome to the session. I am Neha and today I am going to help you with the following question. The question says solve the following linear equation and the equation is x upon 2 minus 1 upon 5 equal to x upon 3 plus 1 upon 4. Let's see the solution. Now we have the equation x upon 2 minus 1 upon 5 equal to x upon 3 plus 1 upon 4. Now to solve these kind of equations we assume that the two sides of the equation that is left hand side and right hand side are balanced. So we perform the same mathematical operations on both sides of the equation so that the balance of this equation is not disturbed. To solve this equation we will multiply both sides of the equation by the Lcm of denominators. So let's see what is the Lcm of denominators. We have denominators 2, 5, 3 and 4. So the Lcm will be, we will first divide all the numbers by 2 that is the common factor is 2. Then again we have one common factor that is 2. Now let's take the common factor 3. Now the last one 5. So the Lcm of denominators is 2 into 2 into 3 into 5 equal to 60. So now let's multiply this equation by 60 or 60 into x upon 2 minus 1 upon 5 equal to 60 into x upon 3 plus 1 upon 4 which implies now to open this bracket first 60 will get multiplied with x upon 2 and then 60 will get multiplied with 1 upon 5. So we get 60 into x upon 2 minus sign as it is 60 into 1 upon 5 equal to again first 60 will get multiplied with x upon 3 and then it will get multiplied with 1 upon 4. So this gives 60 into x upon 3 plus 60 into 1 upon 4 which in turn gives here 2 and 60 will get cancelled with the common factor 2. So 2 times 1 is 2 and 2 times 30 is 16. Here also we will cancel 5 and 60 by the common factor 5. So 5 times 1 is 5 and 5 times 12 is 60. So we will get 30 into x 30 x minus 12 into 1 12 equal to now here 3 and 60 will get cancelled with the common factor 3 and we will get 20 here. Here also we can cancel it with 4 as 4 is a common factor. So 4 times 1 is 4 and 4 times 15 is 60. This gives 20 into x 20 x plus 15 into 1 15. Now we will transpose the variable term 20 x from right-hand side to left-hand side and the constant term 12 from left-hand side to right-hand side. And you know when you transpose one term from left-hand side to right-hand side or from right-hand side to left-hand side then their sign also gets changed. So let's see what it gives. Here the sign of 12 was minus. So when it shifts or when it gets transposed on right-hand side its sign changes to plus. So we get 30 x minus 20 x is 10 x and 30 x minus 20 x is 10 equal to 15 plus 12 is 27. Now we need to find the value of x. So on LHS to get x we will divide both the sides by 10 or 10 x upon 10 equal to 27 upon 10. Here we can give the reason dividing both sides by 10 which implies so 10 and 10 will get cancelled and we will get x here equal to 27 upon 10 as 27 and 10 cannot be factorized further. So this gives us our final answer that is x equal to 27 upon 10. Now let's check whether our solution is correct or not. So take down the equation again that is x upon 2 minus 1 upon 5 equal to x upon 3 plus 1 upon 4. To check whether our solution is correct or not we will substitute the value of x on LHS and on RHS. And we will check whether LHS is equal to RHS or not. So let's take LHS first that is x upon 2 minus 1 upon 5. Now let's substitute the value of x. Now x upon 2 can be written as 1 upon 2 into x minus 1 upon 5 and now we will substitute the value of x that is 1 upon 2 into x is 27 upon 10 minus 1 upon 5 which is equal to 1 into 27 is 27 and 2 into 10 is 20 minus 1 upon 5. Now we will take the LCL LGM of 20 and 5 is 20. So 20 divided by 20 is 1 and 1 into 27 is 27. Now we will substitute the value of x minus 20 divided by 5 is 4 and 4 into 1 is 4 which is equal to 27 minus 4 is 23 upon 20. So our LHS is 23 upon 20. Let's take RHS now that is x upon 3 plus 1 upon 4. Now x upon 3 can be written as x into 1 upon 3 plus 1 upon 4 plus 1 upon 4. Now let's substitute the value of x that is 27 upon 10. So this will be equal to 27 upon 10 into 1 upon 3 plus 1 upon 4 which is equal to 27 into 1 is 27 upon 10 into 3 is 30 plus 1 upon 4. Now let's take the LGM of 30 and 4 that is 60. Now let's take LGM now 60 divided by 30 is 2 and 2 into 27 is 54 plus 16 divided by 4 is 15 and 15 into 1 is 15. So this is equal to 69 over 60. Now they have a common factor 3. So we will cancel them with 3. So numerator gets cancelled with 3 and gives 23 and denominator gets cancelled with 3 and gives 20. So we got 23 upon 20. Thus our LHS is equal to RHS. Therefore we can say the solution x equal to 27 upon 10 is equal to 27 upon 10 is equal to 27 upon 10 is correct. So with this we finished this session. Hope you must have enjoyed it. Goodbye, take care and have a nice day.